Answer to Question #116401 in Mechanics | Relativity for Foibe Kambala

Question

7. A 65 kg trampoline artist jumps vertically upward from the top of platform with speed of
5.0 m/s .
a) How fast is he going as he lands on the trampoline 3.0 m below?
b) If the trampoline behaves like a spring with spring stiffness constant 6.2 x 10 4 N/m.
How far does he depress it?

Answer 1

a)

applying $v^2=u^2+2gS$

$v^2=25+2\times10\times3=95\\v=9.75m/s$

b)

if the trampoline behaves like a spring then the depression in the spring in the spring will create potential energy

so applying energy conservation

$\frac12mv^2=\frac12kx^2$

$65\times(9.75)^2=6.2\times10^4\times x^2$

$x=31.56\times10^{-2}m$

x=31.56cm